Question

Question #0e76f

Answer 1

The answer is `n=29`

Explanation:

To solve this, we can use a very known result, which is the formula to the sum of the `n` first squares. It is as follows:

`\sum_[k=1]^n k^2 = (n(n+1)(2n+1))/6`

Substituting the formula into the equation, we get

`1/(n(2n+1)) * (n(n+1)(2n+1))/6 = 5`

Simplifying:

`(n+1)/6 = 5`

All we need to do now is solve for `n`

`n+1 = 30 => n=29`//

Proof to the formula of the sum of the `n` first squares can be found here: http://mathforum.org/library/drmath/view/56920.html

Hope it helps.